Chapter 2 - Differentiation - Review Exercises - Page 158: 59

$y' = \frac{1-cos(2x)}{2}$

$y = \frac{x}{2} - \frac{sin2x}{4}$ $y = \frac{2x}{4} - \frac{sin(2x)}{4}$ $y = \frac{2x-sin(2x) }{4}$ $y = \frac{1}{4}(2x-sin(2x))$ $y' = (0)(2x-sin(2x)) + \frac{1}{4}(2-2cos(2x))$ $y' = \frac{2-2cos(2x)}{4}$ $y' = \frac{2(1-cos(2x))}{4}$ $y' = \frac{1-cos(2x)}{2}$